A004146 -id:A004146 - cp's OEIS frontend

A005248 Bisection of Lucas numbers: a(n) = L(2n) = A000032(2n).

2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203

Offset: 0

N. J. A. Sloane

Drop initial 2; then iterates of A050411 do not diverge for these starting values. - David W. Wilson
All nonnegative integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)=A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004
a(n+1) = B^(n)AB(1), n>=0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 3=`10`, 7=`010`, 18=`0010`, 47=`00010`, ..., in Wythoff code. a(0) = 2 = B(1) in Wythoff code.
Output of Tesler's formula (as well as that of Lu and Wu) for the number of perfect matchings of an m X n Möbius band where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
Numbers having two 1's in their base-phi representation. - Robert G. Wilson v, Sep 13 2010
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 18 2013: (Start)
a(n) is also one half of the total number of round trips, each of length 2*n, on the graph P_4 (o-o-o-o) (the simple path with 4 points (vertices) and 3 lines (or edges)). See the array and triangle A198632 for the general case of the graph P_N (there N is n and the length is l=2*k).
O.g.f. for w(4,l) (with zeros for odd l): y*(d/dy)S(4,y)/S(4,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form. One half of this o.g.f. for x -> sqrt(x) produces the g.f. (2-3x)/(1-3x+x^2) given below. (End)
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy - 5. - Michel Lagneau, Feb 01 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 7xy + y^2 + 45 = 0. - Colin Barker, Feb 16 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 320 = 0. - Colin Barker, Feb 16 2014
a(n) are the numbers such that a(n)^2-2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
All sequences of this form, b(n+1) = 3*b(n) - b(n-1), regardless of initial values, which includes this sequence, yield this sequence as follows: a(n) = (b(j+n) + b(j-n))/b(j), for any j, except where b(j) = 0. Also note formula below relating this a(n) to all sequences of the form G(n+1) = G(n) + G(n-1).  - Richard R. Forberg, Nov 18 2014
A non-simple continued fraction expansion for F(2n*(k+1))/F(2nk) k>=1 is a(n) + (-1)/(a(n) + (-1)/(a(n) + ... +  (-1)/a(n))) where a(n) appears exactly k times (F(n) denotes the n-th Fibonacci number). E.g., F(16)/F(12) equals 7 + (-1)/(7 + (-1)/7). Furthermore, these a(n) are exactly the positive integers k such that the non-simple infinite continued fraction k + (-1)/(k + (-1)/(k + (-1)/(k + ...))) belongs to Q(sqrt(5)). Compare to Benoit Cloitre and Thomas Baruchel's comments at A002878.  - Greg Dresden, Aug 13 2019
For n >= 1, a(n) is the number of cyclic up-down words of length 2*n over an alphabet of size 3. - Sela Fried, Apr 08 2025

			G.f. = 2 + 3*x + 7*x^2 + 18*x^3 + 47*x^4 + 123*x^5 + 322*x^6 + 843*x^7 + ... - _Michael Somos_, Aug 11 2009

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Richard P. Stanley, Enumerative combinatorics, Vol. 2. Volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard André-Jeannin, Summation of Certain Reciprocal Series Related to Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 200-204.
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Pooja Bhadouria, Deepika Jhala and Bijendra Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Vol. 8, No. 1 (2014), pp. 81-92; sequences B_1, T_1 and R_1.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 51, 56.
Noureddine Chair, Exact two-point resistance, and the simple random walk on the complete graph minus N edges, Ann. Phys., Vol. 327, No. 12 (2012), pp. 3116-3129, Eq. (18).
Tony Crilly, Double sequences of positive integers, Math. Gaz., Vol. 69 (1985), pp. 263-271.
Pedro P. B. de Oliveira, Enrico Formenti, Kévin Perrot, Sara Riva and Eurico L. P. Ruivo, Non-maximal sensitivity to synchronism in periodic elementary cellular automata: exact asymptotic measures, arXiv:2004.07128 [cs.FL], 2020.
Robert G. Donnelly, Molly W. Dunkum, Murray L. Huber and Lee Knupp, Sign-alternating Gibonacci polynomials, arXiv:2012.14993 [math.CO], 2020.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5, No. 15 (2014), pp. 2226-2234.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
Sela Fried, A formula for the number of up-down words, arXiv:2503.02005 [math.CO], 2025.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1986.
Tanya Khovanova, Recursive Sequences.
Emrah Kılıç, Yücel Türker Ulutaş and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq., Vol. 14 (2011), Article 11.5.6, Table 2.
Wentao T. Lu and F. Y. Wu, Close-packed dimers on nonorientable surfaces, Physics Letters A, Vol. 293, No. 5-6 (2002), pp. 235-246. [From Sarah-Marie Belcastro, Jul 04 2009]
Yun-Tak Oh, Hosho Katsura, Hyun-Yong Lee and Jung Hoon Han, Proposal of a spin-one chain model with competing dimer and trimer interactions, arXiv:1709.01344 [cond-mat.str-el], 2017.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Sara Riva, Factorisation of Discrete Dynamical Systems, Ph.D. Thesis, Univ. Côte d'Azur (France 2023).
Jeffrey Shallit, An interesting continued fraction, Math. Mag., Vol. 48, No. 4 (1975), pp. 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., Vol. 48, No. 4 (1975), pp. 207-211. [Annotated scanned copy]
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Jeffrey Shallit and N. J. A. Sloane, Correspondence, 1975.
Paweł J. Szabłowski, On moments of Cantor and related distributions, arXiv preprint arXiv:1403.0386 [math.PR], 2014.
Glenn Tesler, Matchings in Graphs on Non-Orientable Surfaces, Journal of Combinatorial Theory, Series B, Vol. 78, No. 2 (2000), pp. 198-231.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Eric Weisstein's World of Mathematics, Phi Number System.
A. V. Zarelua, On Matrix Analogs of Fermat’s Little Theorem, Mathematical Notes, Vol. 79, No. 6 (2006), pp. 783-796. Translated from Matematicheskie Zametki, Vol. 79, No. 6 (2006), pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A002878 (odd-indexed Lucas numbers), A001906 (Chebyshev S(n-1, 3)), a(n) = sqrt(4+5*A001906(n)^2), A228842.
a(n) = A005592(n)+1 = A004146(n)+2 = A065034(n)-1.
First differences of A002878. Pairwise sums of A001519. First row of array A103997.
Cf. A153415, A201157. Also Lucas(k*n): A000032 (k = 1), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

a005248 n = a005248_list !! n
a005248_list = zipWith (+) (tail a001519_list) a001519_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a:= n-> (<<2|3>>. <<3|1>, <-1|0>>^n)[1$2]: seq(a(n), n=0..30); # Alois P. Heinz, Jul 31 2008
with(combinat): seq(5*fibonacci(n)^2+2*(-1)^n, n= 0..26);

a[0] = 2; a[1] = 3; a[n_] := 3a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Jan 30 2004 *)
Fibonacci[1 + 2n] + 1/2 (-Fibonacci[2n] + LucasL[2n]) (* Tesler. Sarah-Marie Belcastro, Jul 04 2009 *)
LinearRecurrence[{3, -1}, {2, 3}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
LucasL[Range[0,60,2]] (* Harvey P. Dale, Sep 30 2014 *)

{a(n) = fibonacci(2*n + 1) + fibonacci(2*n - 1)}; /* Michael Somos, Jun 23 2002 */

{a(n) = 2 * subst( poltchebi(n), x, 3/2)}; /* Michael Somos, Jun 28 2003 */

[lucas_number2(n,3,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

a(n) = Fibonacci(2*n-1) + Fibonacci(2*n+1).
G.f.: (2-3*x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n, 3) - S(n-2, 3) = 2*T(n, 3/2) with S(n-1, 3) = A001906(n) and S(-2, x) = -1. U(n, x)=S(n, 2*x) and T(n, x) are Chebyshev's U- and T-polynomials.
a(n) = a(k)*a(n - k) - a(n - 2k) for all k, i.e., a(n) = 2*a(n) - a(n) = 3*a(n - 1) - a(n - 2) = 7*a(n - 2) - a(n - 4) = 18*a(n - 3) - a(n - 6) = 47*a(n - 4) - a(n - 8) etc., a(2n) = a(n)^2 - 2. - Henry Bottomley, May 08 2001
a(n) = A060924(n-1, 0) = 3*A001906(n) - 2*A001906(n-1), n >= 1. - Wolfdieter Lang, Apr 26 2001
a(n) ~ phi^(2*n) where phi=(1+sqrt(5))/2. - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(0)=2, a(1)=3, a(n) = 3*a(n-1) - a(n-2) = a(-n). - Michael Somos, Jun 28 2003
a(n) = phi^(2*n) + phi^(-2*n) where phi=(sqrt(5)+1)/2, the golden ratio. E.g., a(4)=47 because phi^(8) + phi^(-8) = 47. - Dennis P. Walsh, Jul 24 2003
With interpolated zeros, trace(A^n)/4, where A is the adjacency matrix of path graph P_4. Binomial transform is then A049680. - Paul Barry, Apr 24 2004
a(n) = (floor((3+sqrt(5))^n) + 1)/2^n. - Lekraj Beedassy, Oct 22 2004
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n (Note: substituting the number 1 for 3 in the last equation gives A000204, substituting 5 for 3 gives A020876). - Creighton Dement, Apr 19 2005
a(n) = (1/(n+1/2))*Sum_{k=0..n} B(2k)*L(2n+1-2k)*binomial(2n+1, 2k) where B(2k) is the (2k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = term (1,1) in the 1 X 2 matrix [2,3] . [3,1; -1,0]^n. - Alois P. Heinz, Jul 31 2008
a(n) = 2*cosh(2*n*psi), where psi=log((1+sqrt(5))/2). - Al Hakanson, Mar 21 2009
From Sarah-Marie Belcastro, Jul 04 2009: (Start)
a(n) - (a(n) - F(2n))/2 - F(2n+1) = 0. (Tesler)
Product_{r=1..n} (1 + 4*(sin((4r-1)*Pi/(4n)))^2). (Lu/Wu) (End)
a(n) = Fibonacci(2n+6) mod Fibonacci(2n+2), n > 1. - Gary Detlefs, Nov 22 2010
a(n) = 5*Fibonacci(n)^2 + 2*(-1)^n. - Gary Detlefs, Nov 22 2010
a(n) = A033888(n)/A001906(n), n > 0. - Gary Detlefs, Dec 26 2010
a(n) = 2^(2*n) * Sum_{k=1..2} (cos(k*Pi/5))^(2*n). - L. Edson Jeffery, Jan 21 2012
From Peter Bala, Jan 04 2013: (Start)
Let F(x) = Product_{n>=0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(3 - sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.31829 56058 81914 31334 ... = 2 + 1/(3 + 1/(7 + 1/(18 + ...))).
Also F(-alpha) = 0.64985 97768 07374 32950 has the continued fraction representation 1 - 1/(3 - 1/(7 - 1/(18 - ...))) and the simple continued fraction expansion 1/(1 + 1/((3-2) + 1/(1 + 1/((7-2) + 1/(1 + 1/((18-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((3^2-4) + 1/(1 + 1/((7^2-4) + 1/(1 + 1/((18^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + 1/2*F(alpha)/F(-alpha) = 1.5142923542... has the simple continued fraction expansion 1 + 1/((3 - 2) + 1/(1 + 1/((18 - 2) + 1/(1 + 1/(123 - 2) + 1/(1 + ...))))). (End)
G.f.: (W(0)+6)/(5*x), where W(k) = 5*x*k + x - 6 + 6*x*(5*k-9)/W(k+1) (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1. Compare with A001906, A002878 and A023039. - Peter Bala, Nov 29 2013
0 = a(n) * a(n+2) - a(n+1)^2 - 5 for all n in Z. - Michael Somos, Aug 24 2014
a(n) = (G(j+2n) + G(j-2n))/G(j), for n >= 0 and any j, positive or negative, except where G(j) = 0, and for any sequence of the form G(n+1) = G(n) + G(n-1) with any initial values for G(0), G(1), including non-integer values. G(n) includes Lucas, Fibonacci. Compare with A081067 for odd number offsets from j. - Richard R. Forberg, Nov 16 2014
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 5*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
From J. M. Bergot, Oct 28 2015: (Start)
For n>0, a(n) = F(n-1) * L(n) + F(2*n+1) - (-1)^n with F(k) = A000045(k).
For n>1, a(n) = F(n+1) * L(n) + F(2*n-1) - (-1)^n.
For n>2, a(n) = 5*F(2*n-3) + 2*L(n-3) * L(n) + 8*(-1)^n. (End)
For n>1, a(n) = L(n-2)*L(n+2) -7*(-1)^n. - J. M. Bergot, Feb 10 2016
a(n) = 6*F(n-1)*L(n-1) - F(2*n-6) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Apr 21 2017
a(n) = F(2*n) + 2*F(n-1)*L(n) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 01 2017
E.g.f.: exp(4*x/(1+sqrt(5))^2) + exp((1/4)*(1+sqrt(5))^2*x). - Stefano Spezia, Aug 13 2019
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(2*n+2) - F(2*n-2) = A001906(n+1) - A001906(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^2 = [1, 1; 1, 2].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 1/a(n) ) = 1/5.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 3/(a(n) + 2/(a(n))) ) = 1/6.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 9/(a(n) + 4/(a(n) + 1/(a(n)))) ) = 1/9.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 3*x^2 + 8*x^3 + 21*x^4 + ... is the o.g.f. for A001906. (End)
a(n) = n + 2 + Sum_{k=1..n-1} k*a(n-k). - Yu Xiao, May 30 2020
Sum_{n>=1} 1/a(n) = A153415. - Amiram Eldar, Nov 11 2020
Sum_{n>=0} 1/(a(n) + 3) = (2*sqrt(5) + 1)/10 (André-Jeannin, 1991). - Amiram Eldar, Jan 23 2022
a(n) = 2*cosh(2*n*arccsch(2)) = 2*cosh(2*n*asinh(1/2)). - Peter Luschny, May 25 2022
a(n) = (5/2)*(Sum_{k=-n..n} binomial(2*n, n+5*k)) - (1/2)*4^n. - Greg Dresden, Jan 05 2023
a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025

Additional comments from Michael Somos, Jun 23 2001

A002315 NSW numbers: a(n) = 6a(n-1) - a(n-2); also a(n)^2 - 2b(n)^2 = -1 with b(n) = A001653(n+1).

Original entry on oeis.org

1, 7, 41, 239, 1393, 8119, 47321, 275807, 1607521, 9369319, 54608393, 318281039, 1855077841, 10812186007, 63018038201, 367296043199, 2140758220993, 12477253282759, 72722761475561, 423859315570607, 2470433131948081, 14398739476117879, 83922003724759193

Offset: 0

N. J. A. Sloane

Named after the Newman-Shanks-Williams reference.
Also numbers k such that A125650(3*k^2) is an odd perfect square. Such numbers 3*k^2 form a bisection of A125651. - Alexander Adamchuk, Nov 30 2006
For positive n, a(n) corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs). - Lekraj Beedassy, Feb 06 2007
Also numbers m such that m^2 is a centered 16-gonal number; or a number of the form 8k(k+1)+1, where k = A053141(m) = {0, 2, 14, 84, 492, 2870, ...}. - Alexander Adamchuk, Apr 21 2007
The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators=A002315 and denominators=A001653. - Clark Kimberling, Aug 27 2008
The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators=A075870, denominators=A002315. - Clark Kimberling, Aug 27 2008
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->oo} a(n) = x*(k*x+1)^n, k = (a(1)-3), x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Numbers k such that (ceiling(sqrt(k*k/2)))^2 = (1+k*k)/2. - Ctibor O. Zizka, Nov 09 2009
A001109(n)/a(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4. - Gary Detlefs, Nov 25 2009
The values 2(a(n)^2+1) are all perfect squares, whose square root is given by A075870. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010
a(n) represents all positive integers K for which 2(K^2+1) is a perfect square. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(8)'s along the main diagonal, and i's along the superdiagonal and subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
Integers k such that A000217(k-2) + A000217(k-1) + A000217(k) + A000217(k+1) is a square (cf. A202391). - Max Alekseyev, Dec 19 2011
Integer square roots of floor(k^2/2 - 1) or A047838. - Richard R. Forberg, Aug 01 2013
Remark: x^2 - 2*y^2 = +2*k^2, with positive k, and X^2 - 2*Y^2 = +2 reduce to the present Pell equation a^2 - 2*b^2 = -1 with x = k*X = 2*k*b and y = k*Y = k*a. (After a proposed solution for k = 3 by Alexander Samokrutov.) - Wolfdieter Lang, Aug 21 2015
If p is an odd prime, a((p-1)/2) == 1 (mod p). - Altug Alkan, Mar 17 2016
a(n)^2 + 1 = 2*b(n)^2, with b(n) = A001653(n), is the necessary and sufficient condition for a(n) to be a number k for which the diagonal of a 1 X k rectangle is an integer multiple of the diagonal of a 1 X 1 square. If squares are laid out thus along one diagonal of a horizontal 1 X a(n) rectangle, from the lower left corner to the upper right, the number of squares is b(n), and there will always be a square whose top corner lies exactly within the top edge of the rectangle. Numbering the squares 1 to b(n) from left to right, the number of the one square that has a corner in the top edge of the rectangle is c(n) = (2*b(n) - a(n) + 1)/2, which is A055997(n). The horizontal component of the corner of the square in the edge of the rectangle is also an integer, namely d(n) = a(n) - b(n), which is A001542(n). - David Pasino, Jun 30 2016
(a(n)^2)-th triangular number is a square; a(n)^2 = A008843(n) is a subsequence of A001108. - Jaroslav Krizek, Aug 05 2016
a(n-1)/A001653(n) is the closest rational approximation of sqrt(2) with a numerator not larger than a(n-1). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. a(n-1)/A001653(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Consider the quadrant of a circle with center (0,0) bounded by the positive x and y axes. Now consider, as the start of a series, the circle contained within this quadrant which kisses both axes and the outer bounding circle. Consider further a succession of circles, each kissing the x-axis, the outer bounding circle, and the previous circle in the series. See Holmes link. The center of the n-th circle in this series is ((A001653(n)*sqrt(2)-1)/a(n-1), (A001653(n)*sqrt(2)-1)/a(n-1)^2), the y-coordinate also being its radius. It follows that a(n-1) is the cotangent of the angle subtended at point (0,0) by the center of the n-th circle in the series with respect to the x-axis. - Graham Holmes, Aug 31 2019
There is a link between the two sequences present at the numerator and at the denominator of the fractions that give the coordinates of the center of the kissing circles. A001653 is the sequence of numbers k such that 2*k^2 - 1 is a square, and here, we have 2*A001653(n)^2 - 1 = a(n-1)^2. - Bernard Schott, Sep 02 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i+4*n+2) - G(i))/(2*G(i+2*n+1)) as long as G(i+2*n+1) != 0. - Klaus Purath, Mar 25 2021
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022
3*a(n-1) is the n-th almost Lucas-cobalancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022
In Moret-Blanc (1881) on page 259 some solution of m^2 - 2n^2 = -1 are listed. The values of m give this sequence, and the values of n give A001653. - Michael Somos, Oct 25 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 6xy + y^2 = 8 = A028884(1). In general, the following applies to all sequences (t) satisfying t(i) = 6t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 6xy + y^2 = A028884(t(1)-6). This includes and interprets the Feb 04 2014 comment on A001541 by Colin Barker as well as the Mar 17 2021 comment on A054489 by John O. Oladokun and the Sep 28 2008 formula on A038723 by Michael Somos. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028884(t(1)-6) always applies.
If (t) is a sequence satisfying t(k) = 7t(k-1) - 7t(k-2) + t(k-3) or t(k) = 6t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) - t(k))/(t(k+n+1) - t(k+n)) always applies, as long as t(k+n+1) - t(k+n) != 0 for integer k and n >= 0. (End)

			G.f. = 1 + 7*x + 41*x^2 + 239*x^3 + 1393*x^4 + 8119*x^5 + 17321*x^6 + ... - _Michael Somos_, Jun 26 2022

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
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Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 247.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

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The Prime Glossary, NSW number.
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Eric Weisstein's World of Mathematics, NSW Number.
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Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Bisection of A001333. Cf. A001109, A001653. A065513(n)=a(n)-1.
First differences of A001108 and A055997. Bisection of A084068 and A088014. Cf. A077444.
Cf. A125650, A125651, A125652.
Row sums of unsigned triangle A127675.
Cf. A053141, A075870. Cf. A000045, A002878, A004146, A026003, A100047, A119915, A192425, A088165 (prime subsequence), A057084 (binomial transform), A108051 (inverse binomial transform).
See comments in A301383.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.
Cf. A000129, A001653, A003499, A358682, A002203.

a002315 n = a002315_list !! n
a002315_list = 1 : 7 : zipWith (-) (map (* 6) (tail a002315_list)) a002315_list
-- Reinhard Zumkeller, Jan 10 2012

I:=[1,7]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A002315 := proc(n)
    option remember;
    if n = 0 then
        1 ;
    elif n = 1 then
        7;
    else
        6*procname(n-1)-procname(n-2) ;
    end if;
end proc: # Zerinvary Lajos, Jul 26 2006, modified R. J. Mathar, Apr 30 2017
a:=n->abs(Im(simplify(ChebyshevT(2*n+1,I)))):seq(a(n),n=0..20); # Leonid Bedratyuk, Dec 17 2017
# third Maple program:
a:= n-> (<<0|1>, <-1|6>>^n. <<1, 7>>)[1, 1]:
seq(a(n), n=0..22);  # Alois P. Heinz, Aug 25 2024

a[0] = 1; a[1] = 7; a[n_] := a[n] = 6a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 20}] (* Robert G. Wilson v, Jun 09 2004 *)
Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {1,7},20]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
Table[ If[n>0, a=b; b=c; c=6b-a, b=-1; c=1], {n, 0, 20}] (* Jean-François Alcover, Oct 19 2012 *)
LinearRecurrence[{6, -1}, {1, 7}, 20] (* Bruno Berselli, Apr 03 2018 *)
a[ n_] := -I*(-1)^n*ChebyshevT[2*n + 1, I]; (* Michael Somos, Jun 26 2022 *)

{a(n) = subst(poltchebi(abs(n+1)) - poltchebi(abs(n)), x, 3)/2};

{a(n) = if(n<0, -a(-1-n), polsym(x^2-2*x-1, 2*n+1)[2*n+2]/2)};

{a(n) = my(w=3+quadgen(32)); imag((1+w)*w^n)};

for (i=1,10000,if(Mod(sigma(i^2+1,2),2)==1,print1(i,",")))

{a(n) = -I*(-1)^n*polchebyshev(2*n+1, 1, I)}; /* Michael Somos, Jun 26 2022 */

a(n) = (1/2)*((1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1)).
a(n) = A001109(n)+A001109(n+1).
a(n) = (1+sqrt(2))/2*(3+sqrt(8))^n+(1-sqrt(2))/2*(3-sqrt(8))^n. - Ralf Stephan, Feb 23 2003
a(n) = sqrt(2*(A001653(n+1))^2-1), n >= 0. [Pell equation a(n)^2 - 2*Pell(2*n+1)^2 = -1. - Wolfdieter Lang, Jul 11 2018]
G.f.: (1 + x)/(1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(n, 6)+S(n-1, 6) = S(2*n, sqrt(8)), S(n, x) = U(n, x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n, 6)= A001109(n+1).
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -8) = a(n). - Benoit Cloitre, Nov 10 2002
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)-b^((2n+1)/2))/2. a(n) = A077444(n)/2. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
a(n) = Sum_{k=0..n} 2^k*binomial(2*n+1, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 08 2003
Same as: i such that sigma(i^2+1, 2) mod 2 = 1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
a(n) = L(n, -6)*(-1)^n, where L is defined as in A108299; see also A001653 for L(n, +6). - Reinhard Zumkeller, Jun 01 2005
a(n) = A001652(n)+A046090(n); e.g., 239=119+120. - Charlie Marion, Nov 20 2003
A001541(n)*a(n+k) = A001652(2n+k) + A001652(k)+1; e.g., 3*1393 = 4069 + 119 + 1; for k > 0, A001541(n+k)*a(n) = A001652(2n+k) - A001652(k-1); e.g., 99*7 = 696 - 3. - Charlie Marion, Mar 17 2003
a(n) = Jacobi_P(n,1/2,-1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
P_{2n}+P_{2n+1} where P_i are the Pell numbers (A000129). Also the square root of the partial sums of Pell numbers: P_{2n}+P_{2n+1} = sqrt(Sum_{i=0..4n+1} P_i) (Santana and Diaz-Barrero, 2006). - David Eppstein, Jan 28 2007
a(n) = 2*A001652(n) + 1 = 2*A046729(n) + (-1)^n. - Lekraj Beedassy, Feb 06 2007
a(n) = sqrt(A001108(2*n+1)). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
a(n) = sqrt(8*A053141(n)*(A053141(n) + 1) + 1). - Alexander Adamchuk, Apr 21 2007
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8), a(1)=1. - Richard Choulet, Sep 18 2007
a(n) = A001333(2*n+1). - Ctibor O. Zizka, Aug 13 2008
a(n) = third binomial transform of 1, 4, 8, 32, 64, 256, 512, ... . - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n) = (-1)^(n-1)*(1/sqrt(-1))*cos((2*n - 1)*arcsin(sqrt(2)). - Artur Jasinski, Feb 17 2010 *WRONG*
a(n+k) = A001541(k)*a(n) + 4*A001109(k)*A001653(n); e.g., 8119 = 17*239 + 4*6*169. - Charlie Marion, Feb 04 2011
In general, a(n+k) = A001541(k)*a(n)) + sqrt(A001108(2k)*(a(n)^2+1)). See Sep 18 2007 entry above. - Charlie Marion, Dec 07 2011
a(n) = floor((1+sqrt(2))^(2n+1))/2. - Thomas Ordowski, Jun 12 2012
(a(2n-1) + a(2n) + 8)/(8*a(n)) = A001653(n). - Ignacio Larrosa Cañestro, Jan 02 2015
(a(2n) + a(2n-1))/a(n) = 2*sqrt(2)*( (1 + sqrt(2))^(4*n) - (1 - sqrt(2))^(4*n))/((1 + sqrt(2))^(2*n+1) + (1 - sqrt(2))^(2*n+1)). [This was my solution to problem 5325, School Science and Mathematics 114 (No. 8, Dec 2014).] - Henry Ricardo, Feb 05 2015
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 7, 0, 41, 0, 239, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -4, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047.
b(n) = 1/2*((-1)^n - 1)*Pell(n) + 1/2*(1 + (-1)^(n+1))*Pell(n+1). The o.g.f. is x*(1 + x^2)/(1 - 6*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*A026003(n-1)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*A026003(n-1)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*Pell(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*Pell(n)*(-x)^n.
Exp( Sum_{n >= 1} 8*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 8*A119915(n)*x^n.
Exp( Sum_{n >= 1} (-8)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 8*A119915(n)*(-x)^n. Cf. A002878, A004146, A113224, and A192425. (End)
E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + cosh(2*sqrt(2)*x))*exp(3*x). - Ilya Gutkovskiy, Jun 30 2016
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^ceiling(k/2). - David Pasino, Jul 09 2016
a(n) = A001541(n) + 2*A001542(n). - A.H.M. Smeets, May 28 2017
a(n+1) = 3*a(n) + 4*b(n), b(n+1) = 2*a(n) + 3*b(n), with b(n)=A001653(n). - Zak Seidov, Jul 13 2017
a(n) = |Im(T(2n-1,i))|, i=sqrt(-1), T(n,x) is the Chebyshev polynomial of the first kind, Im is the imaginary part of a complex number, || is the absolute value. - Leonid Bedratyuk, Dec 17 2017
a(n) = sinh((2*n + 1)*arcsinh(1)). - Bruno Berselli, Apr 03 2018
a(n) = 5*a(n-1) + A003499(n-1), a(0) = 1. - Ivan N. Ianakiev, Aug 09 2019
From Klaus Purath, Mar 25 2021: (Start)
a(n) = A046090(2*n)/A001541(n).
a(n+1)*a(n+2) = a(n)*a(n+3) + 48.
a(n)^2 + a(n+1)^2 = 6*a(n)*a(n+1) + 8.
a(n+1)^2 = a(n)*a(n+2) + 8.
a(n+1) = a(n) + 2*A001541(n+1).
a(n) = 2*A046090(n) - 1. (End)
3*a(n-1) = sqrt(8*b(n)^2 + 8*b(n) - 7), where b(n) = A358682(n). - Stefano Spezia, Nov 26 2022
a(n) = -(-1)^n - 2 + Sum_{i=0..n} A002203(i)^2. - Adam Mohamed, Aug 22 2024
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 3), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
For arbitrary x, a(n+x)^2 - 6*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 8 with a(n) := (1/2)*((1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1)) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(2) * A001542(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/8 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A081554(n) + 1/A081554(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 2*(1 - 1/A055997(k+2))). (End)

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

Original entry on oeis.org

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) =  x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
Alex Fink, Richard K. Guy and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A001110 Square triangular numbers: numbers that are both triangular and square.

Original entry on oeis.org

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100

Offset: 0

N. J. A. Sloane

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004
For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007
The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012
Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012
For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013
For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014
The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014
The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

			a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 204.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-259.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.

Kade Robertson, Table of n, a(n) for n = 0..600 [This replaces an earlier b-file computed by T. D. Noe]
Nikola Adžaga, Andrej Dujella, Dijana Kreso, and Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares, The Mathematical Gazette, 2002, pp. 423-431, esp pp. 424-426.
K. S. Brown, Square Triangular Numbers.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
Yong-Gao Chen and Jin-Hui Fang, Triangular numbers in geometric progression, INTEGERS 7 (2007), #A19.
F. Javier de Vega, On the parabolic partitions of a number, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
Colin Dickson et al., ratio of integers = sqrt(2), thread in newsgroup alt.math.recreational, March 7, 2004.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
Bill Gosper, The Triangular Squares, 2014.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020. Mentions this sequence.
Gillian Hatch, Pythagorean Triples and Triangular Square Numbers, Mathematical Gazette 79 (1995), 51-55.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Roger B. Nelsen, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
A. Nowicki, The numbers a^2+b^2-dc^2, J. Int. Seq. 18 (2015) # 15.2.3.
Matthew Parker, The first 5000 square triangular numbers (7-Zip compressed file).
J. L. Pietenpol, A. V. Sylwester, E. Just, and R. M. Warten, Problem E1473, Amer. Math. Monthly, Vol. 69, No. 2 (Feb. 1962), pp. 168-169. (From the editorial note on p. 169 of this source, we learn that the question about the existence of perfect squares in the sequence of triangular numbers cropped up in the Euler-Goldbach Briefwechsel of 1730; the translation into English of the relevant letters can be found at Correspondence of Leonhard Euler with Christian Goldbach (part II), pp. 614-615.) - _José Hernández_, May 24 2022
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
D. A. Q., Triangular square numbers: a postscript, Math. Gaz., 56 (1972), 311-314.
K. Ramsey, Re_Generalized_Proof_re_Square_Triangular_Numbers. [Broken link]
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Fabio Roman, On certain ratios regarding integer numbers which are both triangulars and squares, arXiv:1703.06701 [math.NT], 2017.
Jon E. Schoenfield, Prime factorization of a(n) for n = 1..300.
Jaap Spies, A Bit of Math, The Art of Problem Solving, Jaap Spies Publishers (2019).
Ralf Stephan, Boring proof of a nonlinearity.
UWC, Problem A, Nieuw Archief voor Wiskunde, Dec 2004; Jaap Spies, Solution.
Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Square triangular number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001108, A001109.
Other S_r type sequences are S_4=A000290, S_5=A004146, S_7=A054493, S_8=A001108, S_9=A049684, S_20=A049683, S_36=this sequence, S_49=A049682, S_144=A004191^2.
Cf. A001014; intersection of A000217 and A000290; A010052(a(n))*A010054(a(n)) = 1.
Cf. A005214, A054686, A232847 and also A233267 (reveals an interesting divisibility pattern for this sequence).
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
See A229131, A182334, A299921 for near-misses.

a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011

[n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015

a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq(A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); # Simon Plouffe in his 1992 dissertation

f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]],AppendTo[lst,f[n]]],{n,0,10!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
Transpose[NestList[Flatten[{Rest[#],34Last[#]-First[#]+2}]&, {0,1},20]][[1]]  (* Harvey P. Dale, Mar 25 2011 *)
LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *)
LinearRecurrence[{6,-1},{0,1},20]^2 (* Harvey P. Dale, Oct 22 2012 *)
(* Square = Triangular = Triangular = A001110 *)
ChebyshevU[#-1,3]^2==Binomial[ChebyshevT[#/2,3]^2,2]==Binomial[(1+ChebyshevT[#,3])/2,2]=={1,36,1225,41616,1413721}[[#]]&@Range[5]
True (* Bill Gosper, Jul 20 2015 *)
L=0;r={};Do[AppendTo[r,L];L=1+17*L+6*Sqrt[L+8*L^2],{i,1,19}];r (* Kebbaj Mohamed Reda, Aug 02 2023 *)

a=vector(100);a[1]=1;a[2]=36;for(n=3,#a,a[n]=34*a[n-1]-a[n-2]+2);a \\ Charles R Greathouse IV, Jul 25 2011

;; With memoizing definec-macro from Antti Karttunen's IntSeq-library.
(definec (A001110 n) (if (< n 2) n (+ 2 (- (* 34 (A001110 (- n 1))) (A001110 (- n 2))))))
;; Antti Karttunen, Dec 06 2013

;; For testing whether n is in this sequence:
(define (inA001110? n) (and (zero? (A068527 n)) (inA001109? (floor->exact (sqrt n)))))
(define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
;; Antti Karttunen, Dec 06 2013

a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001
a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006
a(n) = A001109(n)^2 = A001108(n)*(A001108(n)+1)/2 = (A000129(n)*A001333(n))^2 = (A000129(n)*(A000129(n) + A000129(n-1)))^2. - Henry Bottomley, Apr 19 2000
a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004
From Antonio Alberto Olivares, Nov 07 2003: (Start)
a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
a(n+1) = (17*A029547(n) - A091761(n) - 1)/16. - R. J. Mathar, Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129(2*n)^2/4 = binomial(A001108,2). - Bill Gosper, Jul 28 2008
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008
a(n) = (1/8)*(sinh(2*n*arcsinh(1)))^2. - Artur Jasinski, Feb 10 2010
a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011
a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011
a(2*n+1) = A002315(n)^2*(A002315(n)^2 + 1)/2. - Ivan N. Ianakiev, Oct 10 2012
a(2*n+1) = ((sqrt(t^2 + (t+1)^2))*(2*t+1))^2, where t = (A002315(n) - 1)/2. - Ivan N. Ianakiev, Nov 01 2012
a(2*n) = A001333(2*n)^2 * (A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * (A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013
a(n) = (A001542(n)/2)^2.
From Peter Bala, Apr 03 2014: (Start)
a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = A096979(2*n-1) for n > 0. - Ivan N. Ianakiev, Jun 21 2014
a(n) = (6*sqrt(a(n-1)) - sqrt(a(n-2)))^2. - Arkadiusz Wesolowski, Apr 06 2015
From Daniel Poveda Parrilla, Jul 16 2016 and Sep 21 2016: (Start)
a(n) = A000290(A002965(2*n)*A002965(2*n + 1)) (after Hugh Darwen).
a(n) = A000217(2*(A000129(n))^2 - (A000129(n) mod 2)).
a(n) = A000129(n)^4 + Sum_{k=0..(A000129(n)^2 - (A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers).
a(n) = A002965(2*n)^4 + Sum_{k=A002965(2*n)^2..A002965(2*n)*A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End)
E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016

A092184 Sequence S_6 of the S_r family.

Original entry on oeis.org

0, 1, 6, 25, 96, 361, 1350, 5041, 18816, 70225, 262086, 978121, 3650400, 13623481, 50843526, 189750625, 708158976, 2642885281, 9863382150, 36810643321, 137379191136, 512706121225, 1913445293766, 7141075053841, 26650854921600, 99462344632561, 371198523608646

Offset: 0

Rainer Rosenthal, Apr 03 2004

The r-family of sequences is S_r(n) = 2*(T(n,(r-2)/2) - 1)/(r-4) provided r is not equal to 4 and S_4(n) = n^2 = A000290(n). Here T(n,x) are Chebyshev's polynomials of the first kind. See their coefficient triangle A053120. See also the R. Stephan link for the explicit formula for s_k(n) for k not equal to 4 (Stephan's s_k(n) is identical with S_r(n)).
An integer n is in this sequence iff mutually externally tangent circles with radii n, n+1, n+2 have Soddy circles (i.e., circles tangent to all three) of rational radius. - James R. Buddenhagen, Nov 16 2005
This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. It is the case P1 = 6, P2 = 8, Q = 1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014
a(n) is the block size of the (n-1)-th design in a sequence of multi-set designs with 2 blocks, see A335649. - John P. McSorley, Jun 22 2020

			a(3)=25 because a(1)=1 and a(2)=6 and a(1)*a(3) = 1*25 = (6-1)^2 = (a(2)-1)^2.

Hugo Pfoertner, Table of n, a(n) for n = 0..250
Marco Abrate, Stefano Barbero, Umberto Cerruti, Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Peter Bala, Linear divisibility sequences and Chebyshev polynomials
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
R. Stephan, Boring proof of a nonlinearity
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

See A001110=S_36 for further references to S_r sequences.

Other members of this r-family are: A007877 (r=2), |A078070| (r=3), A004146 (r=5), A054493 (r=7). A098306, A100047. A001353, A001834. A001350, A052530.

[Floor(1/2*(-2+(2+Sqrt(3))^n+(2-Sqrt(3))^n)): n in [0..30]]; // Vincenzo Librandi, Oct 14 2015

A092184 := proc(n)
    option remember;
    if n <= 1 then
        n;
    else
        4*procname(n-1)-procname(n-2)+2 ;
    end if ;
end proc:
seq(A092184(n),n=0..10) ;# Zerinvary Lajos, Mar 09 2008

Table[Simplify[ -((2 + Sqrt[3])^n - 1)*((2 - Sqrt[3])^n - 1)]/2, {n, 0, 26}] (* Stefan Steinerberger, May 15 2007 *)
LinearRecurrence[{5,-5,1},{0,1,6},27] (* Ray Chandler, Jan 27 2014 *)
CoefficientList[Series[x (1 + x)/(1 - 5 x + 5 x^2 - x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 14 2015 *)

Vec(x*(1+x)/(1 - 5*x + 5*x^2 - x^3) + O(x^50)) \\ Michel Marcus, Oct 14 2015

S_r type sequences are defined by a(0)=0, a(1)=1, a(2)=r and a(n-1)*a(n+1) = (a(n)-1)^2. This sequence emanates from r=6.
a(n) = 1/2*(-2 + (2+sqrt(3))^n + (2-sqrt(3))^n). - Ralf Stephan, Apr 14 2004
G.f.: x*(1+x)/(1 - 5*x + 5*x^2 - x^3) = x*(1+x)/((1-x)*(1 - 4*x + x^2)). [from the Ralf Stephan link]
a(n) = T(n, 2)-1 = A001075(n)-1, with Chebyshev's polynomials T(n, 2) of the first kind.
a(n) = b(n) + b(n-1), n >= 1, with b(n):=A061278(n) the partial sums of S(n, 4) = U(n, 2) = A001353(n+1) Chebyshev's polynomials of the second kind.
An integer k is in this sequence iff k is nonnegative and (k^2 + 2*k)/3 is a square. - James R. Buddenhagen, Nov 16 2005
a(0)=0, a(1)=1, a(n+1) = 3 + floor(a(n)*(2+sqrt(3))). - Anton Vrba (antonvrba(AT)yahoo.com), Jan 16 2007
a(n) = 4*a(n-1) - a(n-2) + 2. - Zerinvary Lajos, Mar 09 2008
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001353(n)^2; a(2*n+1) = A001834(n)^2.
a(n) = u(n)^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(6)*u(n-1) - u(n-2) for n >= 2.
Equivalently, a(n) = U(n-1,sqrt(6)/2)^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = (1/2)*( ((sqrt(6) + sqrt(2))/2)^n - ((sqrt(6) - sqrt(2))/2)^n )^2.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -2; 1, 3] and T(n,x) denotes the Chebyshev polynomial of the first kind. Cf. A098306.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
exp( Sum_{n >= 1} 2*a(n)*x^n/n ) = 1 + Sum_{n >= 1} A052530(n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(2*x)*cosh(sqrt(3)*x) - cosh(x) - sinh(x). - Stefano Spezia, Oct 13 2019

Extension and Chebyshev comments from Wolfdieter Lang, Sep 10 2004

A001350 Associated Mersenne numbers.

Original entry on oeis.org

0, 1, 1, 4, 5, 11, 16, 29, 45, 76, 121, 199, 320, 521, 841, 1364, 2205, 3571, 5776, 9349, 15125, 24476, 39601, 64079, 103680, 167761, 271441, 439204, 710645, 1149851, 1860496, 3010349, 4870845, 7881196, 12752041, 20633239, 33385280, 54018521, 87403801, 141422324

Offset: 0

N. J. A. Sloane, R. K. Guy

a(n) is last term in the period of the continued fraction expansion of phi^n (phi being the golden number). E.g.: n=10, phi^10=[122,1,121,1,121,1,121,...] (and the period may only have 1 or 2 terms). Also, a(n) = floor(phi^n)-((n+1) mod 2), or a(n) = A014217(n)-((n+1) mod 2). - Thomas Baruchel, Nov 05 2002 [continued fraction value corrected by Jon E. Schoenfield, Jan 20 2019]
a(n) is the resultant of the polynomials x^2-x-1 and x^(n+1)-x^n-1 for n >= 1. - Richard Choulet, Aug 05 2007
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m). - Michael Somos, Feb 12 2012
Gives the number of arrangements of black and white beads on a necklace with a total of n beads satisfying (1) there is at least one black bead (2) between any two black beads the number of white beads is even and (3) rotations and flippings of a necklace are considered distinct (see Butler). - Peter Bala, Mar 06 2014
This is the case P1 = 1, P2 = 0, Q = -1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
The resultant of the (s_2, s_2+n) pair, where s_n(X) is X^n-X-1, is -a(n). See Rush link. - Michel Marcus, Sep 30 2019

			G.f. = x + x^2 + 4*x^3 + 5*x^4 + 11*x^5 + 16*x^6 + 29*x^7 + 45*x^8 + 76*x^9 + ...
n=1: a(9)/a(3) = 76/4 = 19; a(18)/a(6) = 5776/16 = 361 = 19^2. - _Bob Selcoe_, Jun 01 2014

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Gene Abrams and Gonzalo Aranda Pino, The Leavitt path algebras of generalized Cayley graphs, arXiv preprint arXiv:1310.4735 [math.RA], 2013.
Michael Baake, Joachim Hermisson, and Peter A. B. Pleasants, The torus parametrization of quasiperiodic LI-classes, J. Phys. A 30 (1997), no. 9, 3029-3056.
Michael Baake, John A. G. Roberts, and Alfred Weiss, Periodic orbits of linear endomorphism of the 2-torus and its lattices, arXiv:0808.3489 [math.DS] (2008).
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 21.
Steve Butler, The art of juggling with two balls or a proof for a modular condition of Lucas numbers, arXiv:1004.4293v2 [math.CO], 2010.
James W. Cannon, William J. Floyd, LeeR Lambert, Walter R. Parry and Jessica S. Purcell, Bitwist manifolds and two-bridge knots, arXiv preprint arXiv:1306.4564 [math.GT], 2013-2015.
James W. Cannon, William J. Floyd, LeeR Lambert, Walter R. Parry and Jessica S. Purcell, Bitwist manifolds and two-bridge knots, Pacific Journal of Mathematics 284 (2016), 1-39.
N. Garnier and O. Ramare, Fibonacci numbers and trigonometric identities, Fibonacci Quart. 46/47 (2008/2009), no. 1, 56-61.
R. K. Guy, Letters to N. J. A. Sloane, June-August 1968
Petros Hadjicostas, Cyclic compositions of a positive integer with parts avoiding an arithmetic sequence, Journal of Integer Sequences, 19 (2016), Article 16.8.2.
C. B. Haselgrove, Associated Mersenne numbers, Eureka, 11 (1949), 19-22.
C. B. Haselgrove, Associated Mersenne numbers [Annotated and scanned copy]
G. I. Lehrer and G. B. Segal, Homology stability for classical regular semisimple varieties, Math. Zeit., 236 (2001), 251-290; see Thm. 7.12.
B. Myers, Number of spanning trees in a wheel, IEE Trans. Circuit Theo. 18 (2) (1971) 280-282, Table 1.
Natascha Neumärker, Realizability of Integer Sequences as Differences of Fixed Point Count Sequences, Journal of Integer Sequences, 12 (2009), Article 09.4.5, Example 10.
Aleksandar Petojević, Marjana Gorjanac Ranitović, Dragan Rastovac, and Milinko Mandić, The Golden Ratio, Factorials, and the Lambert W Function, J. Int. Seq. (2024) Art. 24.5.7. See p. 2.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
David E. Rush, Degree n Relatives of the Golden Ratio and Resultants of the Corresponding Polynomials, Fib. Q. 50(4), 2012, 313-325. See p. 317.
Jianxin Wei and Yujun Yang, Associated Mersenne graphs, arXiv:2407.08237 [math.CO], 2024. See p. 4.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).

Cf. A031367, A098554, A000032, A004146, A129744.

[Floor(-(1 - ((1 + Sqrt(5))/2)^n - (-(1 + Sqrt(5))/2)^(-n) + (-1)^n)): n in [0..40]]; // Vincenzo Librandi, Aug 15 2011

A001350 := n -> add(binomial(k-1, 2*k-n)*n/(n-k), k=0..n-1);
seq(A001350(n), n=0..39); # Peter Luschny, Sep 26 2014

Clear[f, n]; f[n_] = -(1 - ((1 + Sqrt[5])/2)^n - (-(1 + Sqrt[5])/2)^(-n) + (-1)^n); Table[FullSimplify[ExpandAll[f[n]]], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Nov 26 2008 *)
a[ n_] := LucasL[n] - 1 - (-1)^n; (* Michael Somos, May 18 2015 *)
a[ n_] := SeriesCoefficient[ x D[ Log[ 1 + x / (1 - x - x^2)], x], {x, 0, n}]; (* Michael Somos, May 18 2015 *)
LinearRecurrence[{1, 2, -1, -1}, {0, 1, 1, 4}, 40] (* Jean-François Alcover, Jan 07 2019 *)

{a(n) = fibonacci(n+1) + fibonacci(n-1) - 1 - (-1)^n};

{a(n) = my(w = quadgen(5)); simplify( -(w^n - 1) * ((-1/w)^n - 1))}; /* Michael Somos, Feb 12 2012 */

from sympy import lucas
def A001350(n): return lucas(n)-((n&1^1)<<1) # Chai Wah Wu, Sep 23 2023

G.f.: x*(1+x^2)/((1-x^2)*(1-x-x^2)). -  Simon Plouffe in his 1992 dissertation
a(n) = a(n-1) + a(n-2) + 1 -(-1)^n. a(-n) = (-1)^n * a(n).
a(n) = A050140(Fibonacci(n)). - Thomas Baruchel, Nov 05 2002
Convolution of F(n) and {1, 0, 2, 0, 2, ...}. a(n) = Sum_{k=0..n} ((1+(-1)^k)-0^k)*F(n-k) = Sum_{k=0..n} F(k)*((1+(-1)^(n-k))-0^(n-k)). - Paul Barry, Jul 19 2004
a(n) = 2*A074331(n) - A000045(n). - Paul Barry, Jul 19 2004
a(n) = Lucas_number(n) - 1 - (-1)^n = A000032(n) - 1 - (-1)^n. - Hieronymus Fischer, Feb 18 2006
a(n) = -(1 - ((1 + sqrt(5))/2)^n - (-(1 + sqrt(5))/2)^(-n) + (-1)^n). - Roger L. Bagula and Gary W. Adamson, Nov 26 2008
a(n) = n * Sum_{k=1..n} (Sum_{i=ceiling((n-k)/2)..(n-k)} (binomial(i,n-k-i)*binomial(k+i-1,k-1))/k*(-1)^(k+1)), n>0. - Vladimir Kruchinin, Sep 03 2010
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4). - Colin Barker, Apr 11 2014
a(n) = sqrt(A152152(n)). - Colin Barker, Apr 11 2014
a(n) = a(2*n)/A000032(n) when n is odd; a(n) = a(2*n)/(A000032(n+2)) when n is even. - Bob Selcoe, Jun 01 2014
a(12n+6)/a(4n+2) = (a(6n+3)/a(2n+1))^2. - Bob Selcoe, Jun 01 2014
a(n) = Sum_{k=0..n-1} binomial(k-1, 2*k-n)*n/(n-k). - Peter Luschny, Sep 26 2014
From Peter Bala, Mar 19 2015: (Start)
a(n) = -(alpha^n - 1)*(beta^n - 1), where alpha = 1/2*(1 + sqrt(5)) and beta = (1/2)*(1 - sqrt(5)).
a(n) = -det(I - M^n) where I is the 2 X 2 identity matrix and M = [ 1, 1; 1, 0 ]. Cf. A129744.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + Sum_{n >= 1} Fibonacci(n)*x^n. Cf. A004146. (End)
a(n) = A052952(n-1) + A052952(n-3). - R. J. Mathar, Jul 02 2018
a(n) = (L(2*n+1) - L(n+1)) mod (L(n+1)-1) for n > 0 where L(k)=A000032(k). - Art Baker, Jan 17 2019
a(n) = Sum_{j=n..2*n-1} L(j) mod Sum_{j=0..n-1} L(j) where L(j)=A000032(j). - Art Baker, Jan 20 2019
Convolution of (1, 0, 3, 0, 5, 0, 7, ...) and (1, 1, 1, 2, 3, 5, 8, 13, ...). - Gary W. Adamson, Jul 08 2019
a(n) = Sum_{d|n} d*A060280(d) = Sum_{d|n} A031367(d). [Baake, Roberts, Weiss, eq(2)]. - R. J. Mathar, Oct 19 2021

Additional comments from Michael Somos, Aug 01 2002

A032198 "CIK" (necklace, indistinct, unlabeled) transform of 1,2,3,4,...

Original entry on oeis.org

1, 3, 6, 13, 25, 58, 121, 283, 646, 1527, 3601, 8678, 20881, 50823, 124054, 304573, 750121, 1855098, 4600201, 11442085, 28527446, 71292603, 178526881, 447919418, 1125750145, 2833906683, 7144450566, 18036423973

Offset: 1

Christian G. Bower

nonn

			From _Petros Hadjicostas_, Jan 07 2018: (Start)
We give some examples to illustrate the theory of C. G. Bower about transforms given in the weblink above. We assume we have boxes of different sizes and colors that we place on a circle to form a necklace. Two boxes of the same size and same color are considered identical (indistinct and unlabeled). We do, however, change the roles of the sequences (a(n): n>=1) and (b(n): n>=1) that appear in the weblink above. We assume (a(n): n>=1) = CIK((b(n): n>=1)).
Since b(1) = 1, b(2) = 2, b(3) = 3, etc., a box that can hold 1 ball only can be of 1 color only, a box that can hold 2 balls only can be one of 2 colors only, a box that can hold 3 balls can be one of 3 colors, and so on.
To prove that a(3) = 6, we consider three cases. In the first case, we have a single box that can hold 3 balls, and thus we have 3 possibilities for the 3 colors the box can be. In the second case, we have a box that can hold 2 balls and a box that can hold 1 ball. Here, we have 2 x 1 = 2 possibilities. In the third case, we have 3 identical boxes, each of which can hold 1 ball. This gives rise to 1 possibility. Hence, a(3) = 3 + 2 + 1 = 6.
To prove that a(4) = 13, we consider 5 cases: a box with 4 balls (4 possibilities), one box with 3 balls and one box with 1 ball (3 possibilities), two identical boxes each with 2 balls (3 possibilities), one box with 2 balls and two identical boxes each with 1 ball (2 possibilities), and four identical boxes each with 1 ball (1 possibility). Thus, a(4) = 4 + 3 + 3 + 2 + 1 = 13.
(End)

Vaclav Kotesovec, Table of n, a(n) for n = 1..1000
A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
C. G. Bower, Transforms (2).
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 31.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Flajolet and M. Soria, The Cycle Construction. [pdf file]
Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
Index entries for sequences related to necklaces

Cf. A000032, A001906, A004146, A005248, A032170, A032287, A088305, A308723, A308747.

Equals A005594(n) - 1.

nmax = 30;
f[x_] = Sum[n*x^n, {n, 1, nmax}];
gf = Sum[(EulerPhi[n]/n)*Log[1/(1 - f[x^n])] + O[x]^nmax, {n, 1, nmax}]/x;
CoefficientList[gf, x] (* Jean-François Alcover, Jul 29 2018, after Joerg Arndt *)

N = 66;  x = 'x + O('x^N);
f(x)=sum(n=1, N, n*x^n );
gf = sum(n=1, N, eulerphi(n)/n*log(1/(1-f(x^n)))  );
v = Vec(gf)
/* Joerg Arndt, Jan 21 2013 */

a(n) = 1/n*Sum_{d divides n} phi(n/d)*A004146(d). - Vladeta Jovovic, Feb 15 2003
From Petros Hadjicostas, Jan 07 2018: (Start)
a(n) = -2 + (1/n)*Sum_{d|n} phi(n/d)*A005248(d) = -2 + (1/n)*Sum_{d|n} phi(n/d)*L(2*d), where L(n) = A000032(n) is the usual Lucas sequence.
G.f.: -Sum_{n >= 1} (phi(n)/n)*log(1 - B(x^n)), where B(x) = x + 2*x^2 + 3*x^3 + 4*x^4 + ... = x/(1-x)^2.
G.f.: -2*x/(1 - x) - Sum_{n>=1} (phi(n)/n)*log(1 - 3*x^n + x^(2*n)).
(End)
From Petros Hadjicostas, Jun 19 2019: (Start)
According to Gibson et al. (2018), a(n) is the number of m-color cyclic compositions of n where each part of size m has m possible colors. This is nothing else than the CIK transform of the sequence 1, 2, 3, 4, ...
Using the theory of Flajolet and Soria (1991), Gibson et al. (2018, Eq. (1.1)) proved that the g.f. of a(n) is Sum_{s >= 1} (phi(s)/s) * log((1 - x^s)^2/(1 - 3*x^s + x^(2*s))), which is exactly the same g.f. as the ones above.
Gibson et al. (2018, p. 3210) also proved that a(n) ~ (2/(3-sqrt(5)))^n/n for large n. See also Chapter 3 in Gibson (2017).
(End)

cp's OEIS Frontend

A275571 Incorrect duplicate of A004146.

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A005248 Bisection of Lucas numbers: a(n) = L(2*n) = A000032(2*n).

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A002315 NSW numbers: a(n) = 6*a(n-1) - a(n-2); also a(n)^2 - 2*b(n)^2 = -1 with b(n) = A001653(n+1).

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A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

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A001110 Square triangular numbers: numbers that are both triangular and square.

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A005248 Bisection of Lucas numbers: a(n) = L(2n) = A000032(2n).

A002315 NSW numbers: a(n) = 6a(n-1) - a(n-2); also a(n)^2 - 2b(n)^2 = -1 with b(n) = A001653(n+1).

A273622 a(n) = (1/3)(Lucas(3n) - Lucas(n)).